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3 years ago

13. Which system of inequalities has no solution?

Mathematics

1 answer:

Anit [1.1K]3 years ago

8 0

Answer:

Option A

Step-by-step explanation:

We have y < 2x - 4 and y > 2x + 1.

=> 2x - 4 > 2x + 1

=> -4 > 1

Since this inequality is false for all real values of x, there is no solution for the system of inequalities.

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Multiplying desimals

kolezko [41]

Answer:

Line up the numbers on the right - do not align the decimal points.

Starting on the right, multiply each digit in the top number by each digit in the bottom number, just as with whole numbers.

Add the products.

Step-by-step explanation:

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2 years ago

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Alexxandr [17]

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Yes

Step-by-step explanation:

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3 years ago

What is the volume of the rectangular prims ?

tangare [24]

They are all different sizes.

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3 years ago

In 2012 your car was worth $10,000. In 2014 your car was worth $8,850. Suppose the value of your car decreased at a constant rat

monitta

Answer:

The function to determine the value of your car (in dollars) in terms of the number of years t since 2012 is:

$f(t) = 10000(0.9407)^t$

Step-by-step explanation:

Value of the car:

Constant rate of change, so the value of the car in t years after 2012 is given by:

$f(t) = f(0)(1-r)^t$

In which f(0) is the initial value and r is the decay rate, as a decimal.

In 2012 your car was worth $10,000.

This means that $f(0) = 10000$ , thus:

$f(t) = 10000(1-r)^t$

2014 your car was worth $8,850.

2014 - 2012 = 2, so:

$f(2) = 8850$

We use this to find 1 - r.

$f(t) = 10000(1-r)^t$

$8850 = 10000(1-r)^2$

$(1-r)^2 = \frac{8850}{10000}$

$(1-r)^2 = 0.885$

$\sqrt{(1-r)^2} = \sqrt{0.885}$

$1 - r = 0.9407$

Thus

$f(t) = 10000(1-r)^t$

$f(t) = 10000(0.9407)^t$

7 0

3 years ago

At an effective annual interest rate of i > 0, each of the following two sets of payments has present value K: (i) A payment

IgorLugansk [536]

Answer:

The present value of K is, $K=251.35$

Step-by-step explanation:

First of all, we need to construct an equation system, so

$(1)K=\frac{169}{(1+i)} +\frac{169}{(1+i)^{2}}$

$(2)K=\frac{225}{(1+i)^{2}} +\frac{225}{(1+i)^{4}}$

Then we equalize both of them so we can find $i$

$(3)\frac{169}{(1+i)} +\frac{169}{(1+i)^{2}}=\frac{225}{(1+i)^{2}} +\frac{225}{(1+i)^{4}}$

To solve it we can multiply $(3)*(1+i)^{4}$ to obtain $(1+i)^{4}*(\frac{169}{(1+i)} +\frac{169}{(1+i)^{2}}=\frac{225}{(1+i)^{2}} +\frac{225}{(1+i)^{4}})$ , then we have $225(1+i)^{2}+225=169(1+i)^{3}+169(1+i)^{2}$ .

This leads to a third-grade polynomial $169i^{3}+451i^{2}+395i-112=0$ , after computing this expression, we find only one real root $i=0.2224$ .

Finally, we replace it in (1) or (2), let's do it in (1) $K=\frac{169}{(1+0.2224)} +\frac{169}{(1+0.2224)^{2}}\\\\K=251.35$

8 0

3 years ago